# Resolution exercises on complex numbers

How do I solve this equation in the field of complex numbers?: $$|z|^2 - z|z| + z = 0$$ My solutions are: $$z_1 = 0$$ $$z_2 = -1$$

• $-1$ is not a solution as you would get $1+1-1\not =0$ – Henry Nov 21 '18 at 11:20

## 5 Answers

$$z=\dfrac{|z|^2}{|z|-1}$$ which is real

If $$z>0,|z|=+z$$ $$0=z^2-z^2+z\iff z=0$$ which is untenable

If $$z\le0,|z|=-z$$ $$0=z^2+z^2+z=z(2z+1)$$

• Your first step only works when $|z| \ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.) – ruakh Nov 21 '18 at 23:29

Observe that $$z=0$$ is a solution of the equation

$$|z|^2 - z|z| + z = 0.$$

( $$z=-1$$ is not a solution !)

Therefore let $$z \ne 0$$ be a further solution of this equation. We get , since $$|z|^2=z \overline{z}:$$

$$\overline{z}-|z|+1=0$$. This gives $$\overline{z}=|z|-1 \in \mathbb R$$.Hence

$$z=|z|-1 \in \mathbb R$$.

If $$z>0$$ , we have $$z=z-1$$, which is impossible. Hence $$z<0$$ and then $$z=-1/2$$.

WLOG $$z=r(\cos t+i\sin t)$$ where $$r>0,t$$ are real

$$r(r-r(\cos t+i\sin t)+(\cos t+i\sin t))=0$$

If $$r\ne0,$$ equating the real & the imaginary parts

$$r-r\cos t+\cos t=0=(1-r)\sin t$$

Case $$\#1:$$

If $$r=1,1=0$$ which is untenable

If $$\sin t=0,$$

Case $$\#2A:\cos t=1,r=0$$ which is untenable

Case $$\#2A:\cos t=-1,r+r-1=0\iff r=?$$

You may also proceed as follows:

• Rewrite the equation to $$|z|^2 - z|z| + z = 0 \Leftrightarrow \boxed{|z|^2 = z(|z|-1)}$$
• Therefore, $$\color{blue}{z}$$ must be $$\color{blue}{\mbox{real}}$$ and so we have $$\color{blue}{|z|^2 = z^2}$$.
• Noting the solution $$\boxed{z = 0}$$ we get $$|z|^2 = z(|z|-1) \stackrel{z \in \mathbb{R}, z \neq 0}{\Leftrightarrow}z = |z|-1 \Rightarrow \boxed{z = -\frac{1}{2}}$$

Let $$r = e^{i\theta}$$.

We get $$r^2 - re^{i\theta}\cdot r + re^{i\theta} = 0$$.

Factorising, we get $$r = 0$$, giving $$z = 0$$ as one solution or:

$$r - re^{i\theta} + e^{i\theta} = 0$$

$$e^{i\theta} = \frac{r}{r-1}$$

From the last equation, the magnitude of the LHS is equal to one. From the RHS, $$e^{i\theta} \in \mathbb{R}$$. Hence $$e^{i\theta} = \pm 1$$.

$$\frac{r}{r-1} = 1$$ gives no solution, but $$\frac{r}{r-1} = -1 \implies r = \frac 12$$. Since $$e^{i\theta} = -1$$ that gives $$z = -\frac 12$$.

Therefore the two solutions are $$z = 0$$ and $$z = -\frac 12$$.